Question: What is the value of the following logarithm? $\log_{9} \left(\dfrac{1}{9}\right)$
Answer: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $9^{y} = \dfrac{1}{9}$ Any number raised to the power $-1$ is its reciprocal, so $9^{-1} = \dfrac{1}{9}$ and thus $\log_{9} \left(\dfrac{1}{9}\right) = -1$.